Chapter 5: Quadratic Functions
Section 5.1: Square Root Property
#1 - 20: Solve the equations using the square root property.
1) 𝑥 2 = 16
2) 𝑦 2 = 25
3) 𝑏 2 = −49
4) 𝑎2 = −16
5) 𝑚2 = 98
6) 𝑑 2 = 24
7) 𝑥 2 = −75
8) 𝑥 2 = −54
9) (𝑥 − 3)2 = 25
10) (𝑥 + 2)2 = 81
11) (2𝑥 − 5)2 = 49
12) (3𝑥 + 7)2 = 121
13) (𝑥 − 4)2 = 150
14) (𝑥 − 8)2 = 48
15) (2𝑥 − 6)2 = −75
16) (5𝑥 + 9)2 = 84
1 2
17) (𝑥 + 3) = 49
1 2
18) (𝑥 − 2) = 16
2 2
19) (𝑥 + 3) = 21
1 2
20) (𝑦 − 5) = 19
#21 - 38: Find a value of C so that the expression becomes a perfect square. Factor your result. We refer to
this method as completing the square.
21) x2 + 6x + C
22) b2 + 8b + C
23) y2 + 10y + C
24) x2 + 16x + C
25) b2 – 4b +C
26) d2 – 12d +C
27) x2 – 14x + C
28) y2 – 20y + C
29) x2 + 5x + C
30) b2 + 7b + C
31) x2 +3x+ C
32) v2 + 9v + C
33) x2 – 7x + C
34) y2 – 5y + C
35) a2 – 11a + C
36) b2 – b + C
1
37) 𝑏 2 + 2 𝑏 + 𝐶
1
38) 𝑎2 − 3 𝑎 + 𝐶
#39 - 62: Solve by completing the square. Specifically, rewrite the equation so it can be solved using the
square root property. That is, first solve for C, like in problems 21-38, and then solve using square roots like
problems 1-24.
39) x2 + 6x = 7
40) b2 + 8b = 9
41) a2 + 10a – 24=0
42) y2 + 6y – 16=0
43) a2 – 10a = 75
44) y2 – 6y = 7
45) x2 – 8x+7 = 0
46) y2 – 4y+3=0
47) x2 + 5x= 6
48) b2 + 3b = 4
49) a2 – 7a –18=0
50) x2 – 5x+24=0
51) x2 + 6x = 5
52) y2 + 10y = 12
53) b2 + 6b = 11
54) x2 – 6x= -4
55) x2 + 8x = -20
56) y2 -6y = -16
57) x2 + 3x + 5 = 0
58) b2 + b – 4 = 0
59) 2x2 + 6x – 5 = 0
60) 3x2 + x – 4 =0
61) 4x2 + x – 3 = 0
62) 2x2 + 23x +11=0
52
Chapter 5: Quadratic Functions
Section 5.2: Quadratic Formula
1) Solve: x2 + 5x – 6 = 0 by
a) Factoring
b) Completing the square
c) The quadratic formula
2)
a)
b)
c)
Solve: x2 + 6x – 7 = 0 by
Factoring
Completing the square
The quadratic formula
3) Solve: y2 + 10y = 5 by
a) Completing the square
b) The quadratic formula
(This can’t be solved by factoring)
4) Solve: b2 + 6b – 3 = 0
a) Completing the square
b) The quadratic formula
(This can’t be solved by factoring)
#5 - 16: Solve using the quadratic formula.
5) y2 + 6y – 16 = 0
6) z2 + 10z – 9 =0
7) x2 + 6x+ 9 = 0
8) d2 + 2d + 1 =0
9) y2 – 2y + 6 = 0
10) x2 – 3x + 5 = 0
11) 2w2 + 3w – 5 =0
12) 3t2 – 7t + 4=0
13) 3z2 – 4z + 3 = 0
14) 2t2 – 6t + 5 = 0
15) 9y2 – 12y + 2 =0
16) 9x2 – 12x + 4 = 0
#17 - 32: Rewrite in the form ax2 + bx + c = 0 with a >0, then solve using the quadratic formula.
17) y2 + 5y = 6
18) x2 + 9x= 10
19) 2t2 + 5t= 8
20) 3x2 – 6x = -2
21) x2 = 3 – 5x
22) y2 = 4 – 5y
23) 3y + 4y2 = 2
24) -5y – 6y2 = 4
25) (x+1)(x – 3)=12
26) (x – 3)(x – 4)= 2
27) 2x(x – 1)= 40
28) 3x(x – 1) = 6
29) 3x(x+1) – 5x = 4
30) 2x(x – 3) +5(x – 4)= 9
31) y(y – 3) + 2y = 4
32) s2 + s(s – 4) = 5s+6
53
Chapter 5: Quadratic Functions
Section 5.3: Graphs of Quadratic Functions
#1 - 12: Make a table of values and sketch the function. Identify the vertex.
1) f(x) = x2
2) g(x) = x2 + 2
3) k(x) = x2 + 4
4) h(x) = x2 – 3
5) n(x) = x2 – 6
6) f(x) = x2 – 2
7) f(x) = 3x2 – 4
8) g(x) = 2x2 – 6
9) b(x) = -5x2 + 12
10) g(x) = -4x2 + 9
11) m(x) = -2x2 +1
12) k(x) = -6x2 + 15
#13 - 24: Make a table of values and sketch the function. Identify the vertex and axis of symmetry.
13) f(x) = (x– 2)2
14) g(x) = (x – 3)2
15) m(x) = (x – 1)2
16) k(x) = (x – 7)2
17) n(x) = (x+3)2
18) h(x) = (x+5)2
19) f(x) = 3(x – 2)2
20) g(x) = 4(x – 1)2
21) b(x) =
−1
(x – 5)2
2
22) f(x) =
−1
(x – 1)2
3
2
23) r(x) = 5 (𝑥 + 1)2
1
24) f(x)= 4(x+2)2
#25 - 36: Make a table of values and sketch the function. Identify the vertex and axis of symmetry.
Identify whether the vertex is a maximum or minimum point, then state the maximum or minimum
value.
25) f(x) = (x – 3)2 + 4
26) g(x) = (x – 2)2 + 6
28) m(x)= 3(x+1)2 + 2
29) g(x) = 2 (𝑥 + 4)2 − 6
30) 𝑓(𝑥) = 3 (𝑥 + 3)2 − 1
31) m(x) = -2x2 + 3
32) t(x) = 3x2 + 6
33) 𝑓(𝑥) = − 4 (𝑥 + 5)2 − 2
34) n(x) = -3(x – 3)2 – 3
35) b(x) = 2(x+3)2 + 4
36) f(x) = -2(x+1)2 + 5
1
27) h(x) = 2(x+3)2 – 4
1
1
#37 – 48: Find the vertex using the vertex formula. Make a table of values and sketch the function.
Identify the vertex and axis of symmetry. Identify whether the vertex is a maximum or minimum point,
then state the maximum or minimum value.
37) f(x) x2 + 6x + 5
38) g(x) = x2 + 10x – 11
39) k(x) =x2 – 4x + 2
40) m(x) = x2 – 2x + 6
41) f(x) = 2x2 + 6x – 3
42) h(x) = -4x2 + 24x – 6
43) f(x) = -x2 + 6x + 4
44) g(x) = x2 – 3x – 2
45) k(x) = -3x2 + 5x – 7
46) g(x) = 2x2 + 7x + 3
47) f(x) = 3x2 – 2x + 1
48) n(x) = -2x2 + 6x + 3
54
Chapter 5: Quadratic Functions
Section 5.4: Applications of Quadratic Functions
1) When a ball is thrown straight upward into the air, the equation h= -16t2 + 80t gives the height (h) in
feet that the ball is above the ground t seconds after it is thrown.
a) How long does it take for the ball to hit the ground?
b) When does the ball reach its maximum height?
c) What is the maximum height of the ball?
2) A student tosses a water balloon of the roof of an apartment building. The height (h) in feet of the
balloon, (t) seconds after it is launched, is given by the formula h = -16t2 +48t + 112. After how many
seconds will the balloon hit the ground?
3) A baby drops his bottle at the peak of a ferris wheel. The height (h) in feet of the bottle, t seconds
after the baby drops the bottle is given by h=-16t2 + 64t + 80. After how many seconds will the bottle
hit the ground?
4) A diver jumps off a cliff to water that is 160 feet below. The diver’s height (h) in feet (t) seconds
after diving is given by h = -16t2 + 160. How long does the dive last?
5) An object fired vertically into the air with an initial velocity of vo feet per second will be at a height
(h) in feet, t seconds after launching, determined by the equation h=vot – 16t2. If the initial velocity is 96
feet per second, at which times will the object have a height of 80 feet?
6) Use the formula in exercise 5 to determine how long it will take the object to return to the ground.
7) Use the formula in exercise 5 to determine the maximum height the object reaches.
8) Harry as 48ft of fencing for use to fence of a rectangular area behind his house for his dog Oswald.
He is going to use the house as one of the sides, thus only has to build three sides out of his fence. What
are the dimensions of the largest area he can fence in?
9) A farmer wishes to fence in a rectangular piece of ground along a river bank for grazing his cattle. If
he has 400 feet of fencing, what should be the dimensions of the rectangle to fence in the greatest area,
using the river as one side of the rectangle?
10) Answer question 8, but assume Harry is not going to use the house for one of the sides of the
rectangular area, and that he needs to construct all 4 sides.
11) Answer question 9, but assume the farmer is not going to use the river bank for one of the sides of
the area he is fencing in, and that he needs to construct all 4 sides.
12) A baker makes a profit of P dollars when he sells n cakes in a day. The baker’s financial analyst
has determined the formula P(n) = -(n – 50)2 + 500 can be used to calculate the daily profit, or of the
baker. What is the optimal number of cakes to sell in a day? What is the maximum profit?
13) The height (in feet) of the water level in a reservoir over a 1-year period is modeled by the function
H(t) = 3.3(t – 9)2 + 14, where t = 1 represents January, t = 2 represents February, and so on. How low
did the water level get that year, and when did it reach its low mark?
55
Chapter 5: Quadratic Functions
Section 5.5: Equations in Quadratic Form
#1 - 6: Solve by isolating the term with the square root, and squaring both sides of the equation to eliminate the
square root. Make sure to check your answers.
1) √𝑥 = 𝑥 − 2
2) √𝑦 = 𝑦 − 12
3) 𝑦 + 2√𝑦 − 15 = 0
4) 2𝑦 + 5√𝑦 − 3 = 0
5) 𝑥 + 2√𝑥 − 8 = 0
6) 𝑏 − 3√𝑏 + 2 = 0
7a) Solve u2 + 5u + 6 = 0
7b) Solve (x-3)2 + 5(x-3) + 6 = 0
8a) Solve u2 + 5u + 4 = 0
8b) Solve (y+7)2 + 5(y+7) + 4 = 0
9a) Solve 2u2 + 5u – 7 = 0
9b) Solve 2(x+4)2 + 5(x+4) – 7 = 0
10a) Solve 3u2 + 13u + 10 = 0
10b) Solve 3(x-6)2 + 13(x– 6) + 10 = 0
#11 - 19: Solve by substitution. That is create and solve a problem with a “u”, then use the solutions to the “u”
problem to solve the given problem.
11) (x+3)2 + 5(x+3)-6= 0
12) (x-5)2 +6(x-5) – 7 = 0
13) (2y+5)2 + 6(2y+5) + 5 = 0
14) 3(x-1)2 + 4(x-1) + 1 = 0
15) 5(x-6)2 + 6(x-6) + 3 = 0
16) (x2 – 1)2 + 5(x2 – 1) + 4 = 0
17) (x2 – 4)2 + 3(x2 – 4) – 4 = 0
18) (
𝑥+2 2
19) (
5
𝑥−3 2
2
𝑥−3
) + 5(
2
)+6 =0
𝑥+2
) + 2(
5
)−3 = 0
#20 - 29: Solve using substitution.
20) x4 + 3x2 – 4 = 0
21) y4 + 8y2 – 9 = 0
23) a4 + 5a2 – 6 = 0
24)
26) 𝑚2⁄3 + 3𝑚1⁄3 − 4 = 0
x-2 +7x-1 – 8 = 0
27) 𝑥 2⁄3 + 5𝑥 1⁄3 − 6 = 0
22) n4 + 3n2 – 24 = 0
25) y-2 + 26y-1 – 27 = 0
28) 𝑏 2⁄5 − 3𝑏 1⁄5 + 2 = 0
29) 𝑥 2⁄5 − 4𝑥 1⁄5 + 3 = 0
56
Chapter 5: Quadratic Functions
Chapter 5: Review
1.
a)
b)
c)
d)
Solve by the square root property.
y2 = 4
k2 – 7 = 0
(q+3)2 = 4
(t+5)2 = -18
2. Solve the quadratic equation by completing the square and applying the square root property.
a) m2 + 6m + 8 = 0
b) p2 + 4p + 6 = 0
3.
a)
b)
c)
d)
Solve the equation using the quadratic formula.
5b2 – 14b – 3 = 0
2t2 + 3t – 7 = 0
12x2 – 5x = 2
w2 + 6w + 14 = 0
4.
a)
b)
c)
d)
Solve
𝑝 + 3√𝑝 = 4
2(5x+3)2 – (5x + 3) – 28 = 0
(x2 + x)2 – 8(x2 + x) = -12
x4 + 5x2 + 4 = 0
5.
a)
b)
c)
Find the vertex and axis of symmetry then sketch a graph of the parabola.
y = (x–2)2 + 3
y = -(x-2)2 – 4
y = -(x + 2)2 + 2
6. Find the vertex by using the vertex formula.
a) T(x) = x2 – 8x +17
b) s(x) = 2x2 – 12x + 19
7. Find the vertex, state if the vertex is a maximum or minimum point, state the maximum or
minimum value, and sketch a graph.
a) y = x2 + 7x + 10
b) y = -2x2 – 12x – 19
57
Chapter 5: Quadratic Functions
Chapter 5: Review
8) The distance (d) an object falls (ignoring air resistance) is given by d = vot + 16t2, where d is
in feet, t is in seconds, and vo is the initial velocity. How long will it take an object to fall 204
feet if it has an initial velocity of 20 feet per second?
9) The depth (in feet) of the snow at the base of a mountain over a 1-year period is modeled by
the function
H(t) = 3.3(t-10)2 + 2, where t = 1 represents January, t = 2 represents February, and so on. How
low did the snow level get that year, and when did it reach its low mark?
10) Harry as 48ft of fencing for use to fence of a rectangular area behind his house for his dog
Oswald. He is going to use the house as one of the sides, thus only has to build three sides out of
his fence. What are the dimensions of the largest area he can fence in?
58